Let's break down and address each question: --- ## **Question 6 [10 marks]** Solve the integro-differential initial value problem: \[ y' - 4 \int_0^t \cos(2\tau - 2t) y(\tau) d\tau = 3e^{3t}, \quad y(0) = 1 \] ### **Step 1: Take Laplace Transform** Let \( Y(s) = \mathcal{L}\{y(t)\} \). Recall: - \( \mathcal{L}\{y'(t)\} = sY(s) - y(0) \) - \( \mathcal{L}\left\{ \int_0^t f(\tau,t) y(\tau) d\tau \right\} = \mathcal{L}\{f(\tau, t)\} * Y(s) \) (convolution) But let's work on the integral term: \[ \int_0^t \cos(2\tau - 2t) y(\tau) d\tau \] Let \( \theta = \tau \), \( d\theta = d\tau \): But \(\cos(2\tau - 2t) = \cos[2(\tau-t)]\). So, \[ \int_0^t \cos(2(\tau-t)) y(\tau) d\tau \] Recall, \( \mathcal{L}\{f(t-\tau)g(\tau)\} = F(s)G(s) \). \[ \mathcal{L}\left\{ \int_0^t \cos(2(\tau-t)) y(\tau) d\tau \right\} = \mathcal{L}\{\cos(-2(t-\tau))\} Y(s) \] But \( \cos(-2(t-\tau)) = \cos(2(t-\tau)) \). So, its Laplace transform is: \[ \mathcal{L}\{\cos(2t)\} = \frac{s}{s^2 + 4} \] So the Laplace of the integral is: \[ \frac{s}{s^2 + 4} Y(s) \] ### **Step 2: Apply Laplace Transform** \[ \mathcal{L}\{y'\} - 4 \mathcal{L}\left\{\int_0^t \cos(2(\tau-t)) y(\tau) d\tau \right\} = \mathcal{L}\{3e^{3t}\} \] \[ (sY(s) - y(0)) - 4\left(\frac{s}{s^2 + 4} Y(s)\right) = \frac{3}{s-3} \] Plug \( y(0) = 1 \): \[ sY(s) - 1 - 4\left(\frac{s}{s^2 + 4} Y(s)\right) = \frac{3}{s-3} \] \[ \left[s - \frac{4s}{s^2 + 4}\right] Y(s) = \frac{3}{s-3} + 1 \] \[ Y(s) \left[\frac{s(s^2+4) - 4s}{s^2+4}\right] = \frac{3}{s-3} + 1 \] \[ Y(s) \left[\frac{s^3 + 4s - 4s}{s^2 + 4}\right] = \frac{3}{s-3} + 1 \] \[ Y(s) \left[\frac{s^3}{s^2 + 4}\right] = \frac{3}{s-3} + 1 \] \[ Y(s) = \left(\frac{3}{s-3} + 1\right) \frac{s^2 + 4}{s^3} \] \[ Y(s) = \frac{3(s^2 + 4)}{s^3(s-3)} + \frac{s^2 + 4}{s^3} \] You can further decompose into partial fractions and then take the inverse Laplace transform to get \( y(t) \). --- ## **Question 7 [10 marks]** A 4 kg mass is attached to a spring with stiffness 64 N/m. Initially, the mass is displaced to a position 2 m to the right and released. No friction or external force. The equation: \( m\ddot{x} + kx = 0 \) Given: - \( m = 4 \) - \( k = 64 \) - \( x(0) = 2 \), \( \dot{x}(0) = 0 \) So, \[ 4\ddot{x} + 64x = 0 \implies \ddot{x} + 16x = 0 \] Solution: \[ x(t) = A\cos(4t) + B\sin(4t) \] Initial conditions: - \( x(0) = 2 \implies A = 2 \) - \( \dot{x}(0) = -4A\sin(0) + 4B\cos(0) = 4B \implies 4B = 0 \implies B = 0 \) So, \[ x(t) = 2\cos(4t) \] **(a) Position after \( \pi/4 \) sec:** \[ x\left(\frac{\pi}{4}\right) = 2\cos\left(4 \cdot \frac{\pi}{4}\right) = 2\cos(\pi) = 2(-1) = -2 \text{ m} \] **(b) Amplitude of oscillatory motion:** - The amplitude is \( 2 \) m. **(c) When does mass cross equilibrium for first time?** - \( x(t) = 0 \implies 2\cos(4t) = 0 \implies \cos(4t) = 0 \) - \( 4t = \frac{\pi}{2} \implies t = \frac{\pi}{8} \) sec --- ### **Summary of Answers** **Question 6:** Laplace method gives \[ Y(s) = \frac{3(s^2 + 4)}{s^3(s-3)} + \frac{s^2 + 4}{s^3} \] Inverse Laplace for \( y(t) \) as described above. **Question 7:** (a) Position after \( \pi/4 \) sec: **-2 m** (b) Amplitude: **2 m** (c) First equilibrium crossing: **\( t = \frac{\pi}{8} \) sec**